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Let $A = \begin{pmatrix} 1 & 1 & 4 \\ -2 & -1 & -2 \\ 5 & 3 & 8 \\ \end{pmatrix} $

Denote W = {B belongs to all the matrices of field $\mathbb{R}^{3\times 3}$ : AB = $ 0_{3\times 3}$}

(a) Prove that W is a linear subspace of all the matrices of field $\mathbb{R}^{3\times 3}$

(b) Find a basis for W.

Thanks!

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    Note that $B$ is in $W$ if and only if every column of $B$ is in $O(A)$ (the nullspace of $A$). – 2017-01-19
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    What did you try to solve this problem? Where do you get stuck? – 2017-01-19

1 Answers 1

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Part (a) is not hard. Clearly $0_{3\times 3} \in W$ and:

  • if $B,C \in W$, then $B+C \in W$ since $A(B+C) = AB+AC = 0_{3\times 3}+0_{3\times 3}=0_{3\times 3}$;
  • if $B \in W$ and $k \in \mathbb{R}$, then $kB \in W$ since $A(kB) = kAB = k0_{3\times 3}=0_{3\times 3}$.

$W$ is thus a linear subspace of the $3 \times 3$-matrices.

For (b), following the hint of Omnomnomnom in his comment, note that if $B = \left( B_1 \vert B_2 \vert B_3 \right)$: $$AB = 0_{3\times 3} \iff A\left( B_1 \vert B_2 \vert B_3 \right) = 0_{3\times 3} \iff \left( AB_1 \vert AB_2 \vert AB_3 \right) = 0_{3\times 3}$$ Each of the columns of $B$ is an element of the null space (or kernel) of $A$. Find this null space and use it to fill the columns of $B$.

Addendum after comments. You found the null space of $A$: $$\mbox{Null}\,A = \left\{ \begin{pmatrix} 2r \\ -6r \\ r \end{pmatrix} : r \in \mathbb{R} \right\}$$ Now construct $B$ by filling its columns with (in general, different) vectors from $\mbox{Null}\,A$: $$B = \begin{pmatrix} 2r_1 & 2r_2 & 2r_3 \\ -6r_1 & -6r_2 & -6r_3 \\ r_1 & r_2 & r_3 \end{pmatrix}$$ This means $B$, and thus any element of $W$, can be written as: $$r_1\begin{pmatrix} 2 & 0 & 0 \\ -6 & 0 & 0 \\ 1 & 0 & 0 \end{pmatrix}+r_2\begin{pmatrix} 0 & 2 & 0 \\ 0& -6 & 0 \\ 0 & 1 & 0 \end{pmatrix}+r_3\begin{pmatrix} 0& 0 & 2 \\ 0 & 0 & -6 \\ 0 & 0&1\end{pmatrix}$$

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    I've found O(A) by row reduction of A. The result was: This system has infinitely many solutions, which are (with parameters ri): { x = 2 r1, y = -6 r1, z = r1 }. I believe this is O(A). Now, how do I use this information to fill the columns? – 2017-01-19
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    Agreed; so the null space of $A$ is one-dimensional. But the columns of $B$ need not be the same; for every column you can pick a (different) element from the null space (e.g. with your notation, use parameter $r_i$ in column $i$). – 2017-01-19
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    I appreciate the hints here. Me and a group of people got to the conclusion that the basis is {2, -6, 1}. However, we cannot understand why it is so. (Or whether this is even true). How do we generate 3 X 3 matrices from that basis. Or maybe we just have the wrong understanding of the basis concept? – 2017-01-19
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    That's a basis for the null space of $A$, not for $W$ (the space of $B$'s). I extended my answer a bit. – 2017-01-19
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    Ah ok. I see it now. Thank you! – 2017-01-19
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    Alright; you're welcome! – 2017-01-19